Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $t \neq 0$. $r = \dfrac{-7t + 63}{-5t^2 - 35t} \times \dfrac{t^2 + 3t - 28}{t - 9} $
Explanation: First factor the quadratic. $r = \dfrac{-7t + 63}{-5t^2 - 35t} \times \dfrac{(t + 7)(t - 4)}{t - 9} $ Then factor out any other terms. $r = \dfrac{-7(t - 9)}{-5t(t + 7)} \times \dfrac{(t + 7)(t - 4)}{t - 9} $ Then multiply the two numerators and multiply the two denominators. $r = \dfrac{ -7(t - 9) \times (t + 7)(t - 4) } { -5t(t + 7) \times (t - 9) } $ $r = \dfrac{ -7(t - 9)(t + 7)(t - 4)}{ -5t(t + 7)(t - 9)} $ Notice that $(t - 9)$ and $(t + 7)$ appear in both the numerator and denominator so we can cancel them. $r = \dfrac{ -7\cancel{(t - 9)}(t + 7)(t - 4)}{ -5t\cancel{(t + 7)}(t - 9)} $ We are dividing by $t + 7$ , so $t + 7 \neq 0$ Therefore, $t \neq -7$ $r = \dfrac{ -7\cancel{(t - 9)}\cancel{(t + 7)}(t - 4)}{ -5t\cancel{(t + 7)}\cancel{(t - 9)}} $ We are dividing by $t - 9$ , so $t - 9 \neq 0$ Therefore, $t \neq 9$ $r = \dfrac{-7(t - 4)}{-5t} $ $r = \dfrac{7(t - 4)}{5t} ; \space t \neq -7 ; \space t \neq 9 $